Question: $\dfrac{3y^2+1}{y^3+y}\cdot\dfrac{dy}{dx}=2x$ and $y(0)=1$. What is $x$ when $y=4$ ? $x=\pm$
Answer: The differential equation is separable. What does it look like after we separate the variables? $\dfrac{3y^2+1}{y^3+y}\,dy=2x\,dx$ Let's integrate both sides of the equation. $\int\dfrac{3y^2+1}{y^3+y}\,dy=\int 2x\,dx$ We can use the substitution $u=y^3+y$ with $du=\left(3y^2+1\right) dy$ on the left side. What do we get? $\ln \left|y^3+y\right|=x^2+C$ What value of $C$ satisfies the initial condition $y(0)=1$ ? Let's substitute $x=0$ and $y=1$ into the equation and solve for $C$. $\begin{aligned} \ln\,\left|1^3+1\right|&=0^2+C\\ \\ C&=\ln 2 \end{aligned}$ Now use this value of $C$ to find $x$ when $y=4$. $\begin{aligned} \ln\,\left|4^3+4\right|&=x^2+\ln 2\\ \\ \\ \ln\left[\left(4^2+1\right)\cdot4\right]&=x^2+\ln 2\\ \\ \\ \ln (17\cdot4)&=x^2+\ln 2\\ \\ x^2 &= \ln (17\cdot4) - \ln 2\\ \\ x^2 &= \ln \dfrac{17\cdot4}2\\ \\ \\ x^2 &= \ln 34\\ \\ x&=\pm \sqrt{\ln 34} \end{aligned}$